正题

P3172

---

题目大意

在 [L,R] 选n个数，问gcd=k的方案数

---

解题思路

因为gcd=k，那么所选的数都是k的倍数，那么可以让L,R整除k，那么有

$$\sum _{a_1=L}^R\sum _{a_2=L}^R...\sum _{a_n=L}^R[gcd(a_1,a_2...a_n)=1]$$

$$\sum _{a_1=l}^r\sum _{a_2=l}^r...\sum _{a_n=l}^r\sum _{d|a_1,d|a_2...d|a_n}\mu (d)$$

$$\sum _{d=1}^n\mu (d)(\frac{r}{d}-\frac{l-1}{d}{)}^n$$

然后可以整除分块，$\mu$ 用杜教筛求

---

code

#include<map>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
#define N 2000210
#define mod 1000000007
using namespace std;
ll n,k,L,R,w,ans,mu[N],p[N],prime[N];
map<ll,ll>smu;
const ll MX=2e6;
void work()
{mu[1]=1;for(ll i=2;i<=MX;++i){if(!p[i]){mu[i]=-1;prime[++w]=i;}for(ll j=1;j<=w&&i*prime[j]<=MX;++j){p[i*prime[j]]=1;if(i%prime[j]==0)break;mu[i*prime[j]]=-mu[i];}}for(ll i=2;i<=MX;++i)mu[i]+=mu[i-1];return;
}
ll S(ll n)
{if(n<=MX)return mu[n];if(smu.find(n)!=smu.end())return smu[n];ll g=1;for(ll l=2,r;l<=n;l=r+1){r=n/(n/l);g-=S(n/l)*(r-l+1);}smu[n]=g;return g;
}
ll ksm(ll x,ll y)
{ll z=1;while(y){if(y&1)z=z*x%mod;x=x*x%mod;y>>=1;}return z;
}
int main()
{work();scanf("%lld%lld%lld%lld",&n,&k,&L,&R);L=(L+k-1)/k-1;R/=k;for(ll l=1,r;l<=R;l=r+1){if(L/l)r=min(L/(L/l),R/(R/l));else r=R/(R/l);(ans+=(S(r)-S(l-1))*ksm(R/l-L/l,n)%mod+mod)%=mod;}printf("%lld",ans);return 0;
}